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\title{Introduction to Quantum Mechanics\\Lecture 9}
\author{Brian Greene\footnote{These lecture notes were TeX'd by Carlos Hernandez, Rohan Bhandari and Steven Castellano}} % ADD YOUR NAME HERE IF YOU WORK ON THIS LECTURE's NOTES
\date{October11, 2011}

\maketitle

\section{Solving the Schr\"{o}dinger Equation}
\subsection{Time Independent Solutions}

This lecture focuses on describing solutions to the Schrodinger Equation (SE). From the previous
lecture (and problem set \#4) we learned that the SE is a separable equation, which can be written as
\begin{equation}
\psi(x) = \phi(t) = C
\end{equation}
where C is a constant.

 Now the SE is no longer a \emph{Partial Differential Equation} (PDE),
but two equivalent \emph{Ordinary Differential Equations} (ODEs), which are much easier to solve. $\phi(t)$ is
then found to be $e^{-wt}$ and $\phi(x)$ depends on what $V(x)$ is. The solution can then be written, as
\begin{equation}
\sum _n C_{n}\psi _n(x)\phi _n(t)
\end{equation} 
which implies that $\Psi(x,t)$ is a summation of wave functions, weighted and normalized by $C_{n}$, with discrete energies determined by a positive integer $n$. 

When $V(x)=0$, $\Psi(x,t)$ can simply be expressed as
\begin{equation}
\Psi (x,t) = \frac{1}{\sqrt{2\pi }}\int _{-\infty }^{\infty }\Phi (k,0)e^{i(\text{kx} - \omega (k)t)}dk = \frac{1}{\sqrt{2\pi }}\int _{-\infty }^{\infty }\Phi (k,0)\psi (x)\phi (t)dk 
\end{equation}
where $\phi(k,0)$ is determined from the initial conditions.

When solving for $V(x) \neq 0$ then the solutions to the time independent SE, $\psi(x)_{n}$, become non-trivial and the general solution turns into the following Fourier series
\begin{equation}
\Psi(x,t) = \sum _n C_{n}\psi _n(x)e^{i\frac{E_{n}t}{\hbar}}
\end{equation}
where $E_{n}$ are the eigenvalues to the corresponding eigenfunctions, $\psi(x)_{n}$. 

Remember, when you have an equation in the form $\hat{H}\psi(x) = E\psi(x)$, where $\hat{H}$ is a linear
operator and E represents a constant, $\hat{H}$ is considered to be a set of eigenfunctions and E, the corresponding set of eigenvalues. So in our case $\hat{H}$ is the spatial part of the SE (or the \emph{Hamiltonian}), which, when applied to a set of $\psi(x)_{n}$, returns eigenvalues, which we will now see are the energy states of the system.

\subsection{E as Energy}

So what allows us to think of $E_{n}$ as energy? Well, let's look at the behavior of the SE in one-dimensional space. 
\begin{equation}
E\psi_{n}(x) = \left(\frac{-\hbar ^2}{2m}\frac{\partial ^2}{\partial x^2} + V(x)\right)\psi_{n}(x) = \frac{-\hbar^2}{2m}\frac{\partial^2\psi_{n}(x)}{\partial x^2}+V(x)\psi_{n}(x) 
\end{equation}
By definition, the second term on the right is the potential energy, $V(x)$,  times $\psi(x)$, where $\psi(x)$ equals $e^{ikx}$. Now, the first term, is a
second derivative of $e^{ikx}$, which is $-k^{2}e^{ikx}$. Now multiply this by $\frac{\hbar^{2}}{2m}$, and substitute $\frac{p}{\hbar}$ for $k$, and you should find that $\frac{p^{2}}{2m}e^{ikx}$. Factor out $\psi(x)$, and you have
\begin{equation}
E=\frac{p_{n}^2}{2m}+V(x) \equiv \text{Kinetic Energy} + \text{Potential Energy} 
\end{equation}
which is just a statement of energy conservation!

Let us now find the expectation value of the total energy in the quantum mechanical system, via our usual way
\begin{equation*}
\langle E \rangle  = \int \psi_{n} ^{*}E (x) \psi_n  dx =  \int \psi_{n}^*\left(\frac{-\hbar ^2}{2m}\frac{\partial ^2}{\partial x^2} + V(x)\right) \psi_n dx = \int \psi_{n} ^{*}E _{n} \psi_{n}  dx
\end{equation*}
where $E_{n}$, again, is just a constant. Pulling this term out of the integral, we are left with
\begin{equation}
\langle E \rangle = E_{n}
\end{equation}
So we have now found that, the case of the time-independent SE, the average value of energy for a given eigenfuntion is just its eigenvalue, $E_n$. During an obversation event, the wave function collapses to a delta function, where only one location is relevant. Similarly, only one $ E_n$ becomes relevant and, therefore, equivalent to $\langle E \rangle$.

One way to quantify the spread of all possible energies is to calculate the variance, $\sigma^2$, of the distribution
\begin{equation}
\sigma ^2 = \left\langle E^2\right\rangle  - \langle E\rangle ^2= \int \psi ^*(E(x) - \langle E\rangle )^2\psi dx
\end{equation}
However, when the wave function collapses, $\langle E\rangle = E(x) = E_{n}$ and we are left with a variance of zero for the delta function, indicating zero spread as expected (please refer to Lecture 5 for more mathematical and graphical details).

\section{An Example: Particle in a Box}
Now, let's do an example using what we've learned. We will start with a canonical example: the Particle in a Box (a.k.a the Infinite Square Well). The potential for such a system is define as the step function 
\begin{equation*}
V(x) = \left\{
\begin{array}{cc}
 0 &  0<x <L \\
 \infty  & \text{Elsewhere}
\end{array}
\right.
\end{equation*}
\begin{figure}[h!]
\centering
    \includegraphics[width=0.35\textwidth]{1-D_Box.jpg}
\caption{An infinite square potential well.}
\end{figure}


 This infinite potential traps the particle within the bounds, as it can never have
enough energy to overcome the box. Our approach to this problem is simple:
\begin{enumerate}[i.]
\item{First, we want to solve the time-independent SE.}
\item{From the time-independent SE, we can learn what the allowed energies of the potential are (remember the
eigenvalue problem from above).}
\item{Finally, we can use the time dependence and initial conditions to determine a general solution.
This will show how the wave function evolves over time.}
\end{enumerate}

Let us begin:
\begin{enumerate}
\item  Note that since there is a zero probability of finding the particle at $x = 0$ or $x = a$, we have
boundary conditions which force $\psi(x=0,a) = 0$. (This is so that {probability equation} = 0 at those
points). Solving the SE (note it is a homogenous second order ODE of the Complex Root case), we find that
{$\psi=$}. Expand the equations into cosine and sine (through Euler's formula) and then group the terms
together: {equations}.

\item Now we need to determine the coefficients $A$ and $B$. We can do this by now using the Boundary conditions.
{equations}.

\item The last step is to normalize the wave function. We can do this because we are normalizing it at $t = 0$,
when $\Psi(x,t) = \psi(x)$. And as proven earlier in the class, if the WF is normalized at $t = 0$, it is normalized for
all time afterwards. {equation including p=}. We now have the time independent wavefunction.
\end{enumerate}
Now to find the allowed energies, one can plug $\psi(x)$ back into Eqn. and calculate the
eigenvalues.

Another approach is that since we have already determined $k$. We know that $k=\frac{p}{\hbar}$ and that $\frac{p^{2}}{2m} = E$.
By just substituting we can find the allowed Energies.
\begin{enumerate}
\item Now remember from the separation of variables, we had $\Psi(x,t)$ = $\psi(x)\phi(t)$. We have just
determined $\psi(x)$. So to get the full WF we just muliply by $\phi(t)$, which we have already determined.
Now we have the full WF and can see how it evolves over time.

\item So our solutions are {eigenfunctions, eigenvalues}={$psi_n$,$E_n$}={equation}
\end{enumerate}

Remarks:

1. {p=} comes in discrete, not continuous, steps.

2. Morever, the wavefunction, $\psi$, is a very particular linear combination of what looks like two free particle wave functions.
This is like a standing wave with 2 momentums. one moving to the right and one to the left. This seems
to imply that $\psi_n$ is no momentum.

a. Does this make sense? Well let's check: ($p=$), so we'll first find $\left\langle x\right\rangle$. Note that
the two factors cancel, and $\left\langle x\right\rangle$ is independent of time, which of course mean $p = m\frac{d\langle x\rangle }{dt} = 0$. So
there are is indeed no net momentum. This is because $\psi_n$ are standing waves, where $e^{ikx}$ and $e^{-
ikx}$ describe the wave moving both to the right and to the left, which cancel and give no net momentum.

3. Note that the wave functions do depend on time. They are not static. Rather their waveform doesn't
change---they arestationary states.

4. One can prove that any $\psi(x)$ can be written as ($\psi(x)=$) for suitable $C_n$.

a. This is true for any well-behaved $V(x)$

5. This can be done for a particle in a box.
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